Jawab:
Nilai stasioner = {8, -8}

Penjelasan dengan langkah-langkah:
Diketahui
f(x) = -8sin(2x + 180°)
f(x) = -8(sin(2x)cos(180°) + sin(180°)cos(2x))
f(x) = -8(sin(2x)(-1) + 0cos(2x))
f(x) = -8(sin(2x)(-1))
--> f(x) = 8(sin(2x))

Periodisitas sin(x) = 2π
Periodisitas sin(2x) = 2π/2
Periodisitas sin(2x) = π
Periodisitas sin(2x) = 180°
dengan (0 < x < 180°) maka
ada 2 nilai stasioner.

Cari turunan f(x)
f(x) = 8(sin(2x))
f'(x) = 8(sin(2x))'
f'(x) = 8 sin'(2x) (2x)'
f'(x) = 8 cos(2x) (2)
f'(x) = 16 cos(2x)

Cari pembuat 0 dari f'(x)
0 = 16 cos(2x)
0 = cos(2x)
2x = arccos(0)
2x = π/2
x  = π/4, x = π - π/4
x₁ = π/4, x₂ = 3π/4

Cari nilai stasionernya
f(x) = 8(sin(2x))
f(x₁) = 8(sin(2x₁))     , f(x₂) = 8(sin(2x₂))
f(π/4) = 8(sin(2π/4)), f(3π/4) = 8(sin(2(3π/4)))
f(π/4) = 8(sin(π/2))  , f(3π/4) = 8(sin(3π/2))
f(π/4) = 8(sin(π/2))  , f(3π/4) = 8(sin(π + π/2))
f(π/4) = 8(sin(π/2))  , f(3π/4) = 8(-sin (π/2))
f(π/4) = 8(1)             , f(3π/4) = 8(-1)
f(π/4) = 8                , f(3π/4) = -8

Nilai stasioner = {8, -8}
(xcvi)